Hi,
I have been trying to build R with optimized BLAS library.
I am using a Ubuntu 13.10 x86_64 desktop, on which I am able to build R
with openblas without any problem:
#BEGIN_SRC sh
./configure --enable-BLAS-shlib --enable-R-shlib LIBnn=lib --disable-nls
--with-blas="-L/usr/lib/openblas-base/
Hi
There is a print method for ecdf
So print(f) should give you an idea of what is going on
See ?ecdf
HTH
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
-Original Message-
From: r-help-boun.
> it gives 'NA' (for whatever y value).
What 'y' values were you using? inf_f maps probabilities (in [0,1]) to
values in the range of the orginal data, x, but it will have problems for
a probability below 1/length(x) because the original data didn't tell
you anything about the ecdf in that region
Thank you, Barradas. It works when finding y, but when I tried to find x
using interpolation for a known y it gives 'NA' (for whatever y value). I
couldn't find out the reason. Any help is really appreciated.
Thanks,
Mano
On Thu, Oct 31, 2013 at 10:53 PM, Rui Barradas wrote:
> Hello,
>
> As fo
Thanks all.
I will "get real" and try to reduce the size of covariance matrix.
Taka
On Fri, Nov 1, 2013 at 8:01 AM, Rolf Turner wrote:
> On 10/31/13 23:14, Takatsugu Kobayashi wrote:
>
>> Hi RUsers,
>>
>> I am struggling to come up with an efficient vectorized way to convert
>> 20Kx20K covar
On 10/31/13 23:14, Takatsugu Kobayashi wrote:
Hi RUsers,
I am struggling to come up with an efficient vectorized way to convert
20Kx20K covariance matrix to a Euclidian distance matrix as a surrogate for
dissimilarity matrix. Hopefully I can apply multidimensional scaling for
mapping these 20K p
Hi Richard
If you cannot get a better suggestion this example from Deepayan Sarkar may
help.
It is way back in the archives and I do not have a reference for it.
I have used it about a year ago as a template to do a complicated key
fl <- grid.layout(nrow = 2, ncol = 6,
height
On Oct 31, 2013, at 1:27 PM, Andreas Leha wrote:
> Hi all,
>
> what is the recommended way to quickly (and without much burden on the
> memory) extract the response from a formula?
If you want its expression value its just form[[2]]
If you wnat it evaluated in the environment of a dataframe th
Hello,
As for the problem of finding y given the ecdf and x, it's very easy,
just use the ecdf:
f <- ecdf(rnorm(100))
x <- rnorm(10)
y <- f(x)
If you want to get the x corresponding to given y, use linear interpolation.
inv_ecdf <- function(f){
x <- environment(f)$x
y <- env
Also, you could try:
library(sqldf)
sqldf('SELECT * FROM df1 EXCEPT SELECT * FROM df2')
A.K.
On Thursday, October 31, 2013 4:28 PM, Rui Barradas
wrote:
Hello,
Try the following. (I don't remember who wrote this function but I saw
it in R-Help)
setdiffDF <- function(A, B){
f <- func
You could e.g. take the data.table package (every data.table is a data.frame)
and make a join:
dt.x <- data.table(x)
dt.y <- data.table(y)
merge.xy <- x[y, nomatch = 0]
diff.xy <- x[!merge.xy]
On 31 Oct 2013, at 21:41, Yasin Gocgun wrote:
> Thanks. Actually, I forgot to add that both have th
Thanks. Actually, I forgot to add that both have the same number of columns.
On Thu, Oct 31, 2013 at 4:07 PM, Bert Gunter wrote:
> lapply() setdiff() by columns.
>
> Unless you have failed to tell us something, you almost certainly will
> not get a data frame (same number of rows/column) as your
Hi All,
I am having some trouble getting lattice to display the legend names by row
instead of by column (default).
Example:
library(lattice)
set.seed(456846)
data <- matrix(c(1:10) + runif(50), ncol = 5, nrow = 10)
dataset <- data.frame(data = as.vector(data), group = rep(1:5, each = 10),
time
Hi all,
what is the recommended way to quickly (and without much burden on the
memory) extract the response from a formula?
The standard way to extract the response from a formula seems to be via
model.frame() or model.extract(), but that is very memory intensive.
Here is a quick example, that (
Hello,
Try the following. (I don't remember who wrote this function but I saw
it in R-Help)
setdiffDF <- function(A, B){
f <- function(A, B)
A[!duplicated(rbind(B, A))[nrow(B) + 1:nrow(A)], ]
df1 <- f(A, B)
df2 <- f(B, A)
rbind(df1, df2)
}
df1 <- data.frame(A = 1:10,
I am creating a scatterplot with the following code.
pl<-ggplot(df,aes(x=Importance,y=Performance,fill=PBF,size=gapsize))+
geom_point(shape=21,colour="black")+scale_size_area(max_size=pointsizefactor)
points are plotted where the size of the point is related to a metric variable
gapsize
lapply() setdiff() by columns.
Unless you have failed to tell us something, you almost certainly will
not get a data frame (same number of rows/column) as your answer.
-- Bert
On Thu, Oct 31, 2013 at 12:58 PM, Yasin Gocgun wrote:
> Hi,
>
> I have two data frames, say, x and y, where y is a subs
Hi,
I have two data frames, say, x and y, where y is a subset of x. How
can I find the set difference of these two data frames (i.e., x-y)?
Thanks,
--
Yasin Gocgun
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
> None of this checks that runs of zero exist in a group; if they don't, you'll
> get warnings
> and -Inf in the output as max takes maxima of nothing. You can add extra
> checks inside
> the function if that bothers you.
Just adding a second argument, 0, to the call to max() will take care of t
Hi Carlos,
With Bioconductor, this can simply be done with:
library(IRanges)
ID <- Rle(1:3, c(3,2,4))
x <- Rle(c(1,0,0,0,0,1,1,0,1))
groups <- split(x, ID)
idx <- groups == 0
Then:
> max(runLength(idx)[runValue(idx)])
1 2 3
2 2 1
Should be fast even with hundreds of thousands
Solution:
http://learnr.wordpress.com/2009/10/22/ggplot2-two-color-xy-area-combo-chart/
--
View this message in context:
http://r.789695.n4.nabble.com/Moving-averages-shading-two-colours-polygon-tp4679460p4679479.html
Sent from the R help mailing list archive at Nabble.com.
_
> If I apply your function to my test data:
>
> the result is
> 1 2 3
> 2 2 2
>
...
> I think f2 does not return the max of consecutive zeros, but the max of any
> consecutve number... Any idea how to fix this?
The toy example of tapply using f2 does indeed return the maximum run lengths
i
Hi,
May be this helps:
fun1 <- function(dat){
lst1 <- lapply(split(dat,dat$ID),function(y){
rl <- rle(y$x)
data.frame(ID=unique(y$ID),MAXZero=max(rl$lengths[rl$values==0]))
})
do.call(rbind,lst1)
}
fun1(df)
# ID MAXZero
#1 1 2
#2 2 2
#3 3 1
A.K.
On Thursday, O
Hi R users,
I am a new user, still learning basics of R. Is there anyway to extract y
(or x) value for a known x (or y) value from ecdf (empirical cumulative
distribution function) curve?
Thanks in advance.
Mano.
[[alternative HTML version deleted]]
_
I am stuck at the following problem.
I have two moving averages. One is faster than another. After plotting them
I need to shade area between them in green when the faster is above slower
and red where faster is below slower.
Something like this:
http://charts.stocktwits.com/production/original_9
If I apply your function to my test data:
ID <- c(1,1,1,2,2,3,3,3,3)
x <- c(1,0,0,0,0,1,1,0,1)
data <- data.frame(ID=ID,x=x)
rm(ID,x)
f2 <- function(x) {
max( rle(x == 0)$lengths )
}
with(data, tapply(x, ID, f2))
the result is
1 2 3
2 2 2
which is not what I'm aiming for. It should be
1 2 3
> > I am trying to compare two different GAM fits.
> > ...
> > How can I get a p-value out of the anova?
See ?anova.gam and pay attention to the 'test' argument.
S Ellison
***
This email and any attachments are confidential. Any u
Hi Jusper
The function plot is generic; the method used to create the image depends on
the type of the data object with which you are working. So for example calling
plot on an "lm" object actually calls plot.lm.
If you type:
> class(mybrick9)
You will see the class of yo
Not terribly verbose. If you google it there is a cgitb flag you can set
to get verbose python output. It is off by default for deployment as it
is a security hole but it is useful now.
Collin.
On Thu, 31 Oct 2013, Erin Hodgess wrote:
> Hi again:
>
> Here is the web output:
>
> Interna
On Oct 30, 2013, at 7:27 PM, Robert Lynch wrote:
> I am trying to compare two different GAM fits.
> I have something like
> Course.bam20 <-bam(zGrade ~ Rep + ISE + White + Female + Years + AP_TOTAL
> + MATH + HSGPA+ EOP + factor(P7APrior, ordered = FALSE)+s(Yfrm7A,k=20),
> data= Course, na.acti
I think you're right...but thanks for your help and time!
Sincerely,
Erin
On Thu, Oct 31, 2013 at 10:01 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> This really has become a python discussion and is probably better held
> elsewhere, but my hunch is that you give a relative p
Hi everybody,
I am using the apriori() fn in the arules package and am encountered an error.
rules <- apriori(rdayst,parameter = list(support = 0.01, confidence = 0.6))
"You chose a very low absolute support count of 0. You might run of memory."
I assume this is related to the value of .01 spe
Dear All,
I often need to do some work on some data which is publicly available on
the EUROSTAT website.
I saw several ways to download automatically mainly the bulk data from
EUROSTAT to later on postprocess it with R, for instance
http://bit.ly/HrDICj
http://bit.ly/HrDL10
http://bit.ly/HrD
This really has become a python discussion and is probably better held
elsewhere, but my hunch is that you give a relative path and Apache doesn't
start python with the pwd you expect.
Michael
On Oct 31, 2013, at 9:12, Erin Hodgess wrote:
> I checked the error logs, and the error appears on
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Carl Witthoft
> Sent: Thursday, October 31, 2013 12:27 PM
> To: r-help@r-project.org
> Subject: Re: [R] Beginner having issues with making pie charts and
> other graphs
>
> R
You've left out all the critical code where you set graphics parameters (par())
to place four figures on one page. Probably you have reduced the margins so
that there is no room for your labels. Note that xlab is printed but not ylab
and there is no space for it.
--
I checked the error logs, and the error appears on the line in which I open
the bz2.R file. It's definitely there, and I set the permissions to 777 for
experimental permissions.
On Thu, Oct 31, 2013 at 7:04 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
>
>
> On Oct 31, 2013, at
That is not an ENVI file. That's an image file... or something.
If you can post a small, reproducible sample of the data array containing
the dates you wish to manipulate, we may be able to help.
--
View this message in context:
http://r.789695.n4.nabble.com/Read-ENVI-files-and-extract-stats
On Oct 31, 2013, at 1:50, Erin Hodgess wrote:
> Hi again:
>
> Here is the web output:
>
> Internal Server Error
>
> The server encountered an internal error or misconfiguration and was unable
> to complete your request.
So your Python code is raising an exception somewhere, not the apache c
At a guess, don't use colour.
John Kane
Kingston ON Canada
> -Original Message-
> From: dimitri.liakhovit...@gmail.com
> Sent: Wed, 30 Oct 2013 14:11:37 -0400
> To: r-help@r-project.org
> Subject: [R] ggplot2 - how to get rid of bar boarder lines
>
> Hello!
>
> I am using ggplot2:
>
>
Without knowing what the data looks like it is a bit difficult to know. See
?dput on how to supply sample data.
However I think that something like the ggplot2 package would be a good way to
go,
John Kane
Kingston ON Canada
> -Original Message-
> From: j.kipl...@cgiar.org
> Sent: Th
> -Original Message-
> So I want to get the max number of consecutive zeros of variable x for each
> ID. I found rle() to be helpful for this task; so I did:
>
> FUN <- function(x) {
> rles <- rle(x == 0)
> }
> consec <- lapply(split(df[,2],df[,1]), FUN)
You're probably better off wit
Rule Number One: Don't use pie charts.
Rule Number Two: NEVER EVER USE PIE CHARTS!
To learn how to create graphs in R, start with ?plot , ?points, ?lines.
Later on you can investigate ?lattice, ?grid, and ?ggplot2 .
The fact that you're looking 'online' instead of in R's builtin help
mechan
On 13-10-30 5:34 PM, Patrick Rioux wrote:
Hi,
Whenever I try to open R from Emacs, it says :
"apply: Searching for program: permission denied, Rterm"
I have the new ESS with the latest Emacs version and R-3.0.2. Also, when I
open Emacs, it says : "No version of R could be found on your system".
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Takatsugu Kobayashi
>
> I am struggling to come up with an efficient vectorized way to convert
> 20Kx20K covariance matrix to a Euclidian distance matrix as a surrogate for
> di
Dear R-helpers,
I need to count the maximum number of consecutive zero values of a variable
in a dataframe by different groups. My dataframe looks like this:
ID <- c(1,1,1,2,2,3,3,3,3)
x <- c(1,0,0,0,0,1,1,0,1)
df <- data.frame(ID=ID,x=x)
rm(ID,x)
So I want to get the max number of consecutive z
I am back to R after a long absence and have a few questions.
I have connected to a very large Microsoft Access database through RODBC in
hopes that I would be able to do some matrix algebra on the many tables I
have loaded into the database. What I am thinking of doing is to assign
specific colum
on/am 31.10.2013 09:12, Prof Brian Ripley wrote/hat geschrieben:
On 30/10/2013 21:15, William Dunlap wrote:
I have to defer to others for policy declarations like how long
the current format used by load and save should be readable.
You could also ask how long R will last
R can still rea
Hi RUsers,
I am struggling to come up with an efficient vectorized way to convert
20Kx20K covariance matrix to a Euclidian distance matrix as a surrogate for
dissimilarity matrix. Hopefully I can apply multidimensional scaling for
mapping these 20K points (commercial products).
I understand that
On 10/31/2013 08:23 PM, Alaios wrote:
Hi everyone,
I am plotting some legend and I am using the axis(at=..) to specify the place
to plot the marks I want.
My plotted data have ncol(x) so the at places have values that span from 1 to
ncol(x)
there I would like to be able to map values that go
Hi there,
I am trying to run an MCMC on stable isotope data from certain organisms to
determine their dietary habits in the package SIAR.
I have prepared my data according to Inger, R., Jackson, A., Parnell, A.,
Bearhop, S. : SIAR V4 (Stable Isotope Analysis in R) an Ecologist's Guide
(also bette
Alex,
it is a bit unclear what you mean by "remap" etc., but maybe
y0 <- 880e6; y1 <- 1020e6
x0 <- 1; x1 <- ncol(x)
y0 + (x0-1):x1 * (y1 - y0)/(x1 - (x0-1))
gives what you want.
Hth -- Gerrit
On Thu, 31 Oct 2013, Alaios wrote:
Hi everyone,
I am plotting some legend and I am usi
Hi
Reading numeric as factor can have many causes from weird formating to some
nonumeric characters. If you can not clean it when making *.csv file you shall
either adopt reading function by using different options
see ?read.table
or if it does not help, you can either polish your values by so
Hi everyone,
I am plotting some legend and I am using the axis(at=..) to specify the place
to plot the marks I want.
My plotted data have ncol(x) so the at places have values that span from 1 to
ncol(x)
there I would like to be able to map values that go from 880e6 to 1020e6.
so
880e6 rem
Dear Eik,
Thanks for your answer. I think indeed I was not to clear of what I want to
achieve. So let me rephrase:
In case that there are aliased predictors in my model, I will see them via the
alias function:
d <- expand.grid(a = 0:1, b=0:1)
d$c <- (d$a + d$b) %% 2
d$y <- rnorm(4)
d <- withi
On 30/10/2013 21:15, William Dunlap wrote:
I have to defer to others for policy declarations like how long
the current format used by load and save should be readable.
You could also ask how long R will last
R can still read (but not write) save() formats used in the 1990's. We
would ex
On Wed, 30 Oct 2013, Katherine Gobin writes:
> Dear R forum,
>
> Just want to know if there is any function / package in R which will
> calculate Yield to Maturity in R for a given bond?
>
> Regards
>
> Katherine
require("sos")
findFn("yield to maturity")
--
Enrico Schumann
Lucerne, Switzer
On Wed, 30 Oct 2013, sart...@voila.fr wrote:
Hi everyone,
I have a data frame with email addresses in the first column and in the second column a list of times (of different lengths) at which an email was sent from the
user in the first column.
Here is an example of my data:
Email Email_sen
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